Ex 11.1, 5
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
the length of the arc
Length of Arc APB = 𝜽/𝟑𝟔𝟎 × (𝟐𝝅𝒓)
= (60°)/(360°) × 2 × 22/7 × 21
= 1/6 × 2 × 22 × 3
= 22 cm
Ex 11.1, 5
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(ii) area of the sector formed by the arc
Area of sector OAPB = 𝜃/360×𝜋𝑟2
= 𝟔𝟎/𝟑𝟔𝟎 × 𝟐𝟐/𝟕 × 𝟐𝟏 × 𝟐𝟏
= 1/6 × 22/7 × 21 × 21
= 1/6 × 22 × 3 × 21
= 231 cm2
Ex 11.1, 5
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(iii) area of segment formed by the corresponding chord
Area of segment APB
= Area of sector OAPB – Area of ΔOAB
From last part,
Area of sector OAPB = 231 cm2
Finding area of Δ AOB
Area Δ AOB = 1/2 × Base × Height
We draw OM ⊥ AB
∴ ∠ OMB = ∠ OMA = 90°
And, by symmetry
M is the mid-point of AB
∴ BM = AM = 1/2 AB
In right triangle Δ OMA
sin O = (side opposite to angle O)/Hypotenuse
sin 𝟑𝟎° = 𝐀𝑴/𝑨𝑶
1/2=𝐴𝑀/21
21/2 = AM
AM = 𝟐𝟏/𝟐
In right triangle Δ OMA
cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝟑𝟎° = 𝑶𝑴/𝑨𝑶
√3/2=𝑂𝑀/21
√3/2 × 21 = OM
OM = √𝟑/𝟐 × 21
From (1)
AM = 𝟏/𝟐AB
2AM = AB
AB = 2AM
Putting value of AM
AB = 2 × 1/2 × 21
AB = 21cm
Now,
Area of Δ AOB = 1/2 × Base × Height
= 𝟏/𝟐 × AB × OM
= 1/2 × 21 × √3/2 × 21
= (𝟒𝟒𝟏√𝟑)/𝟒 cm2
Therefore,
Area of segment APB = Area of sector OAPB – Area of ΔOAB
= (231 – 𝟒𝟒𝟏/𝟒 √𝟑) cm2

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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